TITLE: About a curious nested radical representation for $\cos 1^\circ$
QUESTION [5 upvotes]: I have found the following nested radical representation.
By using the triple angle formula for the cosine, $\cos 3\theta$, and making $\theta = 1^\circ$, we get the cubic equation $ 4x^3-3x = \cos 3^\circ $. In this step , I'll use a trick that is rarely used, by expressing $ x $ as
$ x = \frac{1}{2}\sqrt{3+\frac{\cos 3^\circ}{x}} $
Now I just iterate for x and in this way I get that
$ \cos 1^\circ= \frac{1}{2}\sqrt{3+\frac{\cos 3^\circ}{\frac{1}{2}\sqrt{3+\frac{\cos 3^\circ}{\frac{1}{2}\sqrt{3+\frac{\cos 3^\circ}{\frac{1}{2}\sqrt{3+...}}}}}}} $
Direct calculation shows that this periodic radical converges to $ \cos 1^\circ$
How to prove rigorously that the left hand side is the limit of the right hand side? That's to say, I need a proof that the nested radical converges to $ \cos 1^\circ $

REPLY [2 votes]: You just have to show that $g:z\mapsto\frac{1}{2}\sqrt{3+\frac{\cos 3^\circ}{z}}$ is a contraction in a neighbourhood of $z=1$ then apply Banach fixed-point theorem. We have:
$$ g'(z)=-\frac{\cos\left(\frac{\pi }{60}\right)}{4 z^2 \sqrt{3+\frac{\cos\left(\frac{\pi }{60}\right)}{z}}}$$
and $|g'(z)|$ is much smaller than $1$ (it is close to $\frac{1}{8}$) in a large neighbourhood of $z=1$. Done.